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Trigonometry - Sine, Cosine, and Tangent ratios

Grade 11A Level

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Sine, Cosine, and Tangent ratios (SOH CAH TOA) define the relationships between the sides and angles of a right-angled triangle. The hypotenuse is the longest side, opposite the 90∘90^{\circ} angle; the opposite side is across from the reference angle θ\theta, and the adjacent side is next to it.

Right-angled triangle labeled with hypotenuse, opposite, and adjacent relative to angle theta.
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The Sine Rule asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B} is used for non-right-angled triangles when we know a side and its opposite angle, plus one other piece of information (AAS or SSA).

Non-right-angled triangle with vertices A, B, C and opposite sides a, b, c.
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The Cosine Rule a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A is used for finding a side when two sides and the included angle are known (SAS), or for finding an angle when three sides are known (SSS).

Triangle highlighting two sides and the included angle for Cosine Rule application.
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Trigonometric graphs for y=sin⁡(x)y = \sin(x), y=cos⁡(x)y = \cos(x), and y=tan⁡(x)y = \tan(x) are periodic. The sine and cosine functions have a period of 360∘360^{\circ} and range between −1-1 and 11.

Graph of y = sin(x) from 0 to 360 degrees.

📐Formulae

sin⁡(θ)=OppositeHypotenuse\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}

cos⁡(θ)=AdjacentHypotenuse\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}

tan⁡(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}

asin⁡A=bsin⁡B=csin⁡C (Sine Rule)\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \text{ (Sine Rule)}

a2=b2+c2−2bccos⁡A (Cosine Rule)a^2 = b^2 + c^2 - 2bc \cos A \text{ (Cosine Rule)}

Area=12absin⁡C\text{Area} = \frac{1}{2}ab \sin C

💡Examples

Problem 1:

In a right-angled triangle ABC, the hypotenuse AC is 12 cm and angle BAC is 35°. Calculate the length of the side BC.

Solution:

BC = 12×sin⁡(35∘)≈6.8812 \times \sin(35^\circ) \approx 6.88 cm

Explanation:

Identify that BC is the opposite side to the given angle and AC is the hypotenuse. Use the sine ratio: sin⁡(35∘)=BC12\sin(35^\circ) = \frac{BC}{12}. Multiply both sides by 12 to solve for BC.

Problem 2:

In triangle PQR, PQ = 7 cm, QR = 10 cm, and PR = 8 cm. Find the size of angle QPR.

Solution:

cos⁡(P)=72+82−1022×7×8=49+64−100112=13112\cos(P) = \frac{7^2 + 8^2 - 10^2}{2 \times 7 \times 8} = \frac{49 + 64 - 100}{112} = \frac{13}{112}; P=cos⁡−1(13112)≈83.3∘P = \cos^{-1}(\frac{13}{112}) \approx 83.3^\circ

Explanation:

Since all three sides of a non-right triangle are known, use the Cosine Rule rearranged for the angle: cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}.

Problem 3:

Calculate the area of a triangle where two sides are 5 cm and 9 cm, and the included angle is 42°.

Solution:

Area = 12×5×9×sin⁡(42∘)≈15.06\frac{1}{2} \times 5 \times 9 \times \sin(42^\circ) \approx 15.06 cm²

Explanation:

Use the formula for the area of a triangle when two sides and the included angle (SAS) are known: 12absin⁡C\frac{1}{2}ab \sin C.

Problem 4:

A ladder 5 m long leans against a vertical wall. The base of the ladder is 3 m from the wall. Calculate the angle θ\theta that the ladder makes with the ground.

Diagram showing a ladder leaning against a wall forming a right-angled triangle.

Solution:

cos⁡(θ)=AdjacentHypotenuse\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} cos⁡(θ)=35\cos(\theta) = \frac{3}{5} θ=cos⁡−1(0.6)\theta = \cos^{-1}(0.6) θ≈53.1∘\theta \approx 53.1^{\circ}

Explanation:

We identify the sides relative to the angle between the ladder and the ground. The ladder length is the hypotenuse (55 m) and the distance from the wall is the adjacent side (33 m). We use the Cosine ratio (CAH).

Problem 5:

In triangle XYZ, XY=12XY = 12 cm, angle ZXY=40∘ZXY = 40^{\circ}, and angle XYZ=75∘XYZ = 75^{\circ}. Calculate the length of side YZ.

Triangle XYZ with given angles 40 and 75 degrees and side XY = 12cm.

Solution:

First find the third angle XZYXZY: Angle Z=180∘−(40∘+75∘)=65∘\text{Angle } Z = 180^{\circ} - (40^{\circ} + 75^{\circ}) = 65^{\circ} Use Sine Rule to find YZYZ (xx): xsin⁡(40∘)=12sin⁡(65∘)\frac{x}{\sin(40^{\circ})} = \frac{12}{\sin(65^{\circ})} x=12×sin⁡(40∘)sin⁡(65∘)x = \frac{12 \times \sin(40^{\circ})}{\sin(65^{\circ})} x≈8.51 cmx \approx 8.51\text{ cm}

Explanation:

Since we have two angles and a side, we first find the angle opposite the known side. Then, we apply the Sine Rule to solve for the unknown side YZYZ.